**Source:**Plus Magazine Issue 55

**Problem:**Let's witness a game. The game involves flipping a coin, which you assume has equal probability of coming up heads or tails. The game is played by two players, A and B, who each select a sequence of three flips. For example, assume that Player A selected "heads-heads-heads" (HHH) and Player B has selected "tails-heads-heads" (THH). Then the coin is flipped repeatedly, resulting in a sequence like the following:

HTHTHHHHTHHHTTTTHTHH...

The player whose sequence showed up first (HHH for Player A or THH for Player B) is declared the winner.

A chooses first and you have to help B. Show that given what A has chosen, B can always choose a sequence such that B is in a better position than A.

For example: If A chooses HHH, B chooses THH to make sure that now it has 7/8 probability to win.

Update (27/07/10):

**Solution:**Posted in comments by me!

The possible sequences that A can select and the particular sequence that B should select to have high probabilty of winning are

ReplyDeleteHHH - THH

HHT - THH

HTH - THT

HTT - THT

THH - HTH

THT - HTH

TTH - HTT

TTT - HTT

To genaralize if A selects a squence like a1a2a3 B should select (-a1)a1a2

@Bhanu: HTH - THT pair seems incorrect. Can you plz. check?

ReplyDelete@Shantanu.. you are correct.. Thanx..

ReplyDeleteP(getting THT before HTH) = 1/2 {Symmetrical terms :P}

@Bhanu.. nice try.. but this would not always work as pointed out by Shantanu..

If you are interested, you can read the complete article here:

http://plus.maths.org/issue55/features/nishiyama/